The lens combination consists of two lenses, | vedclass

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(B) For the first lens $L_1$ (convex lens): Focal length $f_1 = +10 	ext{ cm}$, object distance $u_1 = -30 	ext{ cm}$. Using the lens formula $\frac

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$60 	ext{ cm}$ to the left of the concave lens

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(B) For the first lens $L_1$ (convex lens): Focal length $f_1 = +10 	ext{ cm}$, object distance $u_1 = -30 	ext{ cm}$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$: $\frac{1}{v_1} - \frac{1}{-30} = \frac{1}{10} \implies \frac{1}{v_1} = \frac{1}{10} - \frac{1}{30} = \frac{3-1}{30} = \frac{2}{30} = \frac{1}{15}$. So, $v_1 = +15 	ext{ cm}$. This image acts as a virtual object for the second lens $L_2$. The distance between the lenses is $d = 3 	ext{ cm}$. Object distance for $L_2$ is $u_2 = +(15 - 3) = +12 	ext{ cm}$. For the second lens $L_2$ (concave lens): Focal length $f_2 = -10 	ext{ cm}$, object distance $u_2 = +12 	ext{ cm}$. Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$: $\frac{1}{v_2} - \frac{1}{12} = \frac{1}{-10} \implies \frac{1}{v_2} = \frac{1}{12} - \frac{1}{10} = \frac{5-6}{60} = -\frac{1}{60}$. So, $v_2 = -60 	ext{ cm}$. The negative sign indicates that the final image is formed $60 	ext{ cm}$ to the left of the concave lens $L_2$.

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